The quadratic equations 2x²-x-1=0 and x²+px-5=0 have common integral root, find the value of p
Answers
Answer:
The most obvious answer is p=1, as the equations would be identical in that case. But are there others?
Roots of the first x=[-2±√(4–4p)]/2p. Roots of the second x=[-2±√(4–4p)]/2
[-2±√(4–4p)]/2=[-2±√(4–4p)]/2p. If both of the ± signs are + or if both are –, then the only solution is when 2=2p. That is, p=1.
But, what about if [-2+√(4–4p)]/2=[-2–√(4–4p)]/2p or if [-2–√(4–4p)]/2=[-2+√(4–4p)]/2p?
(1) In the first case the equation simplifies to -1 + √(1–p)=-1–√(1-p)/p
This becomes -p+p√(1–p)=-1–√(1–p)
Solve by isolating radicals and squaring both sides.
p√(1–p)+√(1–p)=p–1
p^2(1–p)+2p(1–p)+(1–p)=p^2–2p+1
p^2–p^3+2p–2p^2+1–p=p^2–2p+1
0=p^3+2p^2–3p
0=p(p^2+2p–3)
0=p(p+3)(p–1)
p=0,1,and -3. 0 is extraneous, but both other solutions work.
(2) In the other case, the equation simplifies to -1 – √(1–p)=-1+√(1-p)/p.
This becomes -p–p√(1-p)=-1+√(1–p)
Solve by isolating radicals and squaring both sides.
p√(1–p)+√(1–p)=1–p
p^2(1–p)+2p(1–p)+(1-p)=1–p
p^2–p^3+2p–2p^2=0
0=p^3+p^2–2p
0=p(p+2)(p–1)
p=-2,0,1. In this case, 0 and -2 are extraneous, only 1 checks.
So, going back to our original problem, we know that 1 works. We need to check -3. Is there a common root for -3x^2+2x+1=0 and x^2+2x–3=0? The first has factors (3x+1)(-x+1)=0, so roots are -1/3 and 1.
The second has factors (x+3)(x–1)=0, so the roots are -3 and 1. Hence there is a common root of 1 for p=-3. In conclusion, there are 2 possible values for p: 1 and -3.
Answer:
2x²-x-1=0
2x²-2x+x-1=0
2x(x-1)+1(x-1)=0
(x-1)(2x+1)=0
x=1 or x= -1/2.
and given common integer root of both equations so, one root of x²+px-5=0 is
x=1.
(x-1)=0
so,the one factor of x²+px-5=0 is (x-1)
and x²+px-5/(x-1) must be divide
so, x-1)x²+px-5(x+(p+1)
x²-x
__-_+__
(p+1)x-5
(p+1)x-(p+1)
_-____+____
p-4=0( definetly)
so, p = 4.
make me brainlist if you like the solution.